Given binary string str of size N, the task is to remove the minimum number of characters from the given binary string such that the characters in the remaining string are in sorted order.
Examples:
Input: str = “1000101”
Output: 2
Explanation: Removal of the first two occurrences of ‘1’ modifies the string to “00001”, which is a sorted order. The string can also be made “00011” by performing 2 removals. Therefore, the minimum count of characters to be removed is 2.
Input: str = “001111”
Output: 0
Explanation: The string is already sorted. Therefore, the minimum count of character to be removed is 0.
Last Occurrence Approach: The optimized approach in linear time and constant space is discussed in Set 1 of this article. Here, we are discussing the Dynamic Programming Approach.
Dynamic Programming Approach: This problem can be solved using dynamic programming by observing the following facts, if K deletion is required to make the string sorted till ith index and
- Case1: S[i+1] = 1 then minimum number of deletions required to make string sorted till (i+1)th index will also be K as appending 1 to a sorted string will keep string sorted so no more deletion required.
- Case2: S[i + 1] = 0 then we have two way to make string sorted till (i+1)th index that are
- either delete all 1’s before (i+1)th index, or
- delete current 0.
Minimum number of deletion to make string valid till (i+1)th index will be minimum of (numbers of 1’s before (i+1)th index , K+1).
Follow the steps below to solve the problem:
- Initialize the variables count1 as 0 and N is the length of the string s.
- Initialize the vector dp[n+1] with values 0.
- Iterate over the range [0, n) using the variable i and perform the following tasks:
- If s[i] equals 0, then set dp[i+1] as the minimum of count1 or 1 + dp[i].
- Else, set dp[i+1] as dp[i] and increase the value of count1 by 1.
- After performing the above steps, print the value of dp[n] as the answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int minDeletions(string s)
{
int n = s.size();
vector<int> dp(n + 1, 0);
int count1 = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
dp[i + 1] = min(count1, 1 + dp[i]);
}
else {
dp[i + 1] = dp[i];
count1++;
}
}
return dp[n];
}
int main()
{
string s = "00101101";
cout << minDeletions(s);
return 0;
}
|
Java
import java.util.*;
public class GFG {
static int minDeletions(String s)
{
int n = s.length();
int[] dp = new int[n + 1];
for (int i = 0; i < n + 1; i++) {
dp[i] = 0;
}
int count1 = 0;
for (int i = 0; i < n; i++) {
if (s.charAt(i) == '0') {
dp[i + 1] = Math.min(count1, 1 + dp[i]);
}
else {
dp[i + 1] = dp[i];
count1++;
}
}
return dp[n];
}
public static void main(String args[])
{
String s = "00101101";
System.out.println(minDeletions(s));
}
}
|
Python3
def minDeletions(s):
n = len(s)
dp = [0] * (n + 1)
count1 = 0
for i in range(n):
if (s[i] == '0'):
dp[i + 1] = min(count1, 1 + dp[i])
else:
dp[i + 1] = dp[i]
count1 += 1
return dp[n]
s = "00101101"
print(minDeletions(s))
|
C#
using System;
public class GFG {
static int minDeletions(string s)
{
int n = s.Length;
int[] dp = new int[n + 1];
for (int i = 0; i < n + 1; i++) {
dp[i] = 0;
}
int count1 = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
dp[i + 1] = Math.Min(count1, 1 + dp[i]);
}
else {
dp[i + 1] = dp[i];
count1++;
}
}
return dp[n];
}
public static void Main()
{
string s = "00101101";
Console.Write(minDeletions(s));
}
}
|
Javascript
<script>
function minDeletions(s) {
let n = s.length;
let dp = new Array(n + 1).fill(0);
let count1 = 0;
for (let i = 0; i < n; i++) {
if (s[i] == '0') {
dp[i + 1] = Math.min(count1,
1 + dp[i]);
}
else {
dp[i + 1] = dp[i];
count1++;
}
}
return dp[n];
}
let s = "00101101";
document.write(minDeletions(s));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach: Space Optimization Approach
In this approach we can improve the space complexity of the algorithm by using two variables to keep track of the minimum and maximum positions of the given element in the array instead of using a 2D vector dp. This will reduce the space complexity from O(N) to O(1).
Steps:
- Declare function minDeletions that take a string as an argument.
- Initialize the variable next and curr to 0 that refers to the current and previous element of dp.
- Now the approach is the same as the previous code but to optimize the space we change dp[i] to curr and dp[i+1] to next because dp[i] is only dependent on dp[i+1] and dp[i].
Implementation :
C++
#include <bits/stdc++.h>
using namespace std;
int minDeletions(string s)
{
int n = s.size();
int next = 0;
int curr = 0;
int count1 = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
next = min(count1, 1 + curr);
}
else {
next = curr;
count1++;
}
curr = next;
}
return next;
}
int main()
{
string s = "00101101";
cout << minDeletions(s);
return 0;
}
|
Python3
def minDeletions(s: str) -> int:
n = len(s)
next, curr, count1 = 0, 0, 0
for i in range(n):
if s[i] == '0':
next = min(count1, 1 + curr)
else:
next = curr
count1 += 1
curr = next
return next
s = "00101101"
print(minDeletions(s))
|
C#
using System;
public class Gfg {
static int minDeletions(string s)
{
int n = s.Length;
int next = 0;
int curr = 0;
int count1 = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '0') {
next = Math.Min(count1, 1 + curr);
}
else {
next = curr;
count1++;
}
curr = next;
}
return next;
}
public static void Main()
{
string s = "00101101";
Console.WriteLine(minDeletions(s));
}
}
|
Javascript
function minDeletions(s) {
let n = s.length;
let next = 0;
let curr = 0;
let count1 = 0;
for (let i = 0; i < n; i++) {
if (s[i] == '0') {
next = Math.min(count1,1 + curr);
} else {
next = curr;
count1++;
}
curr = next;
}
return next;
}
let s = "00101101";
console.log(minDeletions(s));
|
Java
import java.util.*;
public class Main {
public static int minDeletions(String s)
{
int n = s.length();
int next = 0;
int curr = 0;
int count1 = 0;
for (int i = 0; i < n; i++) {
if (s.charAt(i) == '0') {
next = Math.min(count1, 1 + curr);
}
else {
next = curr;
count1++;
}
curr = next;
}
return next;
}
public static void main(String[] args)
{
String s = "00101101";
System.out.println(minDeletions(s));
}
}
|
Time Complexity: O(N)
Auxiliary Space: O(1)